Find remainder when $ x^{3}+4x^{2}-4x+6 $ is divided by $ x-5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&1&4&-4&6\\& & 5& 45& \color{black}{205} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{41}&\color{orangered}{211} \end{array} $$The remainder when $ x^{3}+4x^{2}-4x+6 $ is divided by $ x-5 $ is $ \, \color{red}{ 211 } $.
Explanation
We can find remainder using synthetic division method.
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&4&-4&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 1 }&4&-4&6\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&4&-4&6\\& & \color{blue}{5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 5 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}5&1&\color{orangered}{ 4 }&-4&6\\& & \color{orangered}{5} & & \\ \hline &1&\color{orangered}{9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 9 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&4&-4&6\\& & 5& \color{blue}{45} & \\ \hline &1&\color{blue}{9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 45 } = \color{orangered}{ 41 } $
$$ \begin{array}{c|rrrr}5&1&4&\color{orangered}{ -4 }&6\\& & 5& \color{orangered}{45} & \\ \hline &1&9&\color{orangered}{41}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 41 } = \color{blue}{ 205 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&4&-4&6\\& & 5& 45& \color{blue}{205} \\ \hline &1&9&\color{blue}{41}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 205 } = \color{orangered}{ 211 } $
$$ \begin{array}{c|rrrr}5&1&4&-4&\color{orangered}{ 6 }\\& & 5& 45& \color{orangered}{205} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{41}&\color{orangered}{211} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 211 }\right) $.